$\int^{2}_{0}x^2{\sqrt{1+x^3}}\,dx\, = $
Solution: Strategy Let's first find the indefinite integral $\int x^2{\sqrt{1+x^3}}\,dx\,$. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int x^2{\sqrt{1+x^3}}\,dx\,$, we can use U-substitution. If we let $ {u=1+x^3}$, then ${du=3x^2 \, dx}$ and ${ x^2\,dx=\dfrac{1}{3}\, du}$. So we have: ∫ x 2 1 + x 3 − − − − − √ d x = ∫ 1 + x 3 − − − − − √ ⋅ x 2 d x = ∫ u √ ⋅ 1 3 d u = 1 3 ∫ u √ d u = 1 3 ⋅ 2 3 u 3 2 + C = 2 9 ( 1 + x 3 ) 3 2 + C \begin{aligned}\int x^2{\sqrt{1+x^3}}\,dx\,&=\int {\sqrt{{1+x^3}}}\cdot {x^2\,dx}\,\\\\\\\\ &=\int\sqrt{ u}\,\cdot {\dfrac13\, du}\,\\\\\\\\ &=\dfrac13\int\sqrt{u}\,du\\\\\\\\ &=\dfrac{1}{3}\cdot \dfrac23u\^{\frac{3}{2}}+C\\\\\\\\ &=\dfrac29\left(1+x^3\right)\^{\frac{3}{2}}+C \end{aligned} Evaluating the definite integral Now let's compute the definite integral: ∫ 2 0 x 2 1 + x 3 − − − − − √ d x = 2 9 ( 1 + x 3 ) 3 2 ∣ ∣ ∣ 2 0 = 2 9 ( 27 − 1 ) = 52 9 \begin{aligned}\int^{2}_{0}x^2{\sqrt{1+x^3}}\,dx\,&=\dfrac29\left(1+x^3\right)\^{\frac{3}{2}}\Bigg|^2_0\\\\\\\\ &=\dfrac{2}{9}\left(27-1\right)\\\\\\\\ &=\dfrac{52}9\end{aligned} [Did we have to find the indefinite integral first?] The answer $\int^{2}_{0}x^2{\sqrt{1+x^3}}\,dx\, = \dfrac{52}{9}$